Question

if vector a is the parallel to vector B what is their vector product​

Answer 1

➡ Given:

✏ vector A is the parallel to vector B.

➡ To Find:

✏ Cross-product (vector-product) of vector A and vector B.

➡ Formula:

✏ Cross-product of two vectors is given by

$\: \: \: \: \: \: \: \: \: \: \: \underline{ \boxed{ \bold{ \rm{ \pink{ \vec{A} \times \vec{B} = | \vec{A}| | \vec{B}| sin \theta}}}}}$

✏ Where, $\theta$ = angle between both vectors.

➡ Calculation:

✏ Here angle between both vectors = 0$\degree$

$: \implies \rm \: \vec{A} \times \vec{B} = | \vec{A}| | \vec{B}| sin0 \degree \\ \\ : \implies \rm \: \underline{ \boxed{ \bold{ \rm{ \orange{ \vec{A} \times \vec{B} = 0}}}}} \: \: ( \because \: sin0 \degree = 0)$

Answer 2

Question :

If the Vector A is parallel to Vector B, the What is their Vector product ?

AnsWer :

0.

Explanation:

• Let two Vector,

$\: \: \tt {\vec{A} \: \: and \: \: \vec B \: two \: vector. \: both} \\ \: \: \tt{ parallel \: to \: each \: other.}$

When, Both Vector parallel to each other then the angle between both vector be 0°.

$\rule{200}3$

Formula Use,

$\tt\: \: \: \: \vec{A} \times \vec{B} = |A| |B| \sin \theta. \hat{ n}$

$\star \tt \: \: here \: \hat{n} \: unit \: vector \: which \: \perp \\ \: \: \: \: \tt{to \: \vec A \: and \: \vec B}$

$\tt \: \: \star \: \theta \: show \: angle \: between \: \vec{A} \: and \: \vec{B}.$

Calculation :

$\tt\vec{A} \times \vec{B} = |A| |B| \sin \theta. \hat{n} \\ \: \: \: \: \: \: \: \: \: \: \: \: \tt= 1. \sin0 \degree. \hat{n}\\</p><p> \: \: \: \: \: \: \: \: \: \: \: \:\tt = 0.$

Some Examples,

We have,

$\tt \hat{i} \times \hat{j} = |\hat{i} | |\hat{j} | \sin 90 \degree \hat{n} \\ \tt \: \: \: \: \: \: \: \: \: \: \: = 1 \times 1 \times \hat{n} \\ \: \: \: \: \: \: \: \: \: \: \: = \hat{k}.$

$\tt \hat{j} \times \hat{k} = |\hat{j} | |\hat{k} | \sin 90 \degree \hat{n} \\ \tt \: \: \: \: \: \: \: \: \: \: \: = 1 \times 1 \times \hat{n} \\ \: \: \: \: \: \: \: \: \: \: \: = \hat{i}.$

$\tt \hat{k} \times \hat{i} = |\hat{k} | |\hat{i} | \sin 90 \degree \hat{n} \\ \tt \: \: \: \: \: \: \: \: \: \: \: = 1 \times 1 \times \hat{n} \\ \: \: \: \: \: \: \: \: \: \: \: = \hat{j}.$

$\tt \hat{j} \times \hat{i} = |\hat{j} | |\hat{i} | \sin 90 \degree \hat{n} \\ \tt \: \: \: \: \: \: \: \: \: \: \: = -1 \times 1 \times \hat{n} \\ \: \: \: \: \: \: \: \: \: \: \: = - \hat{k}.$

$\tt \hat{k} \times \hat{j} = |\hat{k} | |\hat{j} | \sin 90 \degree \hat{n} \\ \tt \: \: \: \: \: \: \: \: \: \: \: =- 1 \times 1 \times \hat{n} \\ \: \: \: \: \: \: \: \: \: \: \: = - \hat{i}.$

$\tt \hat{i} \times \hat{k} = |\hat{i} | |\hat{k} | \sin 90 \degree \hat{n} \\ \tt \: \: \: \: \: \: \: \: \: \: \: = -1 \times 1 \times \hat{n} \\ \: \: \: \: \: \: \: \: \: \: \: = - \hat{j}.$

$\tt\star \:\sin0 \degree=0.\\</p><p></p><p>\tt\star\:\sin90\degree=1.$